The Birthday Problem
Becky Schmoyer
It's your birthday! Congratulations! But
wait… doesn't the person sitting
behind you have a birthday today also? What are the odds of
that!?!
The Birthday Problem is a way of finding the
probability of two people in the
same room having the same birthday. To start, say there is one
person. That
person has a possibility of 365 days as being his birthday. (This
is assuming
there are 365 days in the year.) Now, if you have two people, the
second
person has 364 different days that could be his and not shared
with the first
person. If you take it up to three people, the third person has
363 different
days that could be his and not shared with the other two.
Therefore, the
probability of a match being made between three people is this
1 - (365) * (364) * (363) / (365) * (365) * (365) = .0082
To get this, first you must figure out the
probability that no one in the
group has a match. To do this, just find the product of each
distinct day over
the total days
365/365 * 364/365 * 363/365 = .9918
Then, to find the probability that at least two
people share the same
birthday, subtract the previous answer from one.
1 - .9918 = .0082
From this, we can extract an equation that will
aid in finding the
probabilities in much larger groups.
P(A) = (365) * (364) * (363) * … (365 - N + 1) / 365^N
P(AC) 1 - (365) * (364) * (363) * … (365 - N + 1) / 365^N
WHEN
N = Number of people in the group, P(A) =
Probability that no one shares a
birthday, P(AC) = Probability that 2 people share the same
birthday
Now, with these formulas and a basic
understanding of The Birthday Problem,
one can figure out the probability of similar birthdays in bigger
groups, and
also with different circumstances. For instance, how likely is it
that in a
group of six students there will be at least two birthdays in the
same month?
To figure this out, we follow the same steps as
above, but use 12 months in
replace of the 365 days. The first person has a range of twelve.
The next has
11 different months, the third, 10, and so on. So, using the
equation above,
we figure that
(12 * 11* 10 * 9 * 8 * 7) / 12^6 = .2228009
1 - .2228009 = .7772
So, the probability of 2 people having their
birthday in the same month is
greater than 2 students out of 6 having the same their birthday
on the same
day. (The probability of 2 out of 6 is around .04)
January, 1999